New methods for (ϕ, Γ) modules

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1 DOI /s z RESEARCH Open Access New methods for (ϕ, Γ) modules iran S edlaya * *Correspondence: kedlaya@ucsd.edu Department of Mathematics, University of California San Diego, 9500 Gilman Drive #0112, La Jolla, CA 92093, USA Abstract We provide new proofs of two key results of p-adic Hodge theory: the Fontaine- Wintenberger isomorphism between Galois groups in characteristic 0 and characteristic p, and the Cherbonnier Colmez theorem on decompletion of (ϕ, Ŵ)-modules. These proofs are derived from joint work with Liu on relative p-adic Hodge theory, and are closely related to the theory of perfectoid algebras and spaces, as in the work of Scholze. eywords: p-adic Hodge theory, Perfectoid fields, Field of norms equivalence, Witt vectors, (ϕ, Ŵ)-modules, Cherbonnier Colmez theorem Mathematics Subject Classification: Primary 11S20, Secondary 11S15, 13F35 To Robert, forever quelling the rebellious provinces Let p be a prime number. The subject of p-adic Hodge theory concerns the interplay between different objects arising from the cohomology of algebraic varieties over p-adic fields. A good introduction to the subject circa 2010 can be found in the notes of Brinon and Conrad [4]; however, in the subsequent years the subject has been radically altered by the introduction of some new ideas and techniques. While these ideas have their origins in work of this author on relative p-adic Hodge theory [15, 16] and were further developed in joint work with Liu [18, 19], they are most widely known through Scholze s work on the theory and applications of perfectoid algebras [21, 22, 23, 24, 25]. The purpose of this paper is to reinterpret and reprove two classic results of p-adic Hodge theory through the optic of perfectoid algebras (but in a self-contained manner). The first of these results is a theorem of Fontaine and Wintenberger [10] on the relationship between Galois theory in characteristic 0 and characteristic p. Theorem (Fontaine Wintenberger) For µ p the group of all p-power roots of unity in an algebraic closure of Q p, the absolute Galois groups of the fields F p ((π)) and Q p (µ p ) are isomorphic (and even homeomorphic as profinite topological groups). The original proof of Theorem depends in a crucial way on higher ramification theory of local fields, as developed for instance in the book of Serre [27]. This causes difficulties when trying to generalize Theorem 0.0.1, e.g., to local fields with imperfect residue fields. We expose here a new approach to Theorem in which ramification 2015 edlaya. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Page 2 of 31 theory plays no role; one instead makes a careful analysis of rings of Witt vectors over valuation rings. In the process, we obtain a far more general result, in which Q p (µ p ) can be replaced by any sufficiently ramified p-adic field; more precisely, we obtain a functorial (and hence compatible with Galois theory) correspondence between perfectoid fields and perfect fields of characteristic p. This is the tilting correspondence in the sense of [21], which is proved using almost ring theory; our proof here is the somewhat more elementary argument found in [18] (see Remark for further discussion). As a historical note, we remark that we learned the key ideas from this proof from attending Coleman s 1997 Berkeley course Fontaine s theory of the mysterious functor, whose principal content appears in [6]. Our second topic is the description of continuous representations of p-adic Galois groups on Q p -vector spaces (such as might arise from étale cohomology with p-adic coefficients) in terms of (ϕ, Ŵ)-modules. The original description of this form was given by Fontaine [9] in terms of a Cohen ring for a field of formal power series, and is an easy consequence of Theorem Our main focus is the refinement of Fontaine s result by Cherbonnier and Colmez [5], in which the Cohen ring is replaced with a somewhat smaller ring of convergent power series (see Theorem for the precise statement). This refinement is critical to a number of applications of p-adic Hodge theory, notably Colmez s construction of the p-adic Langlands correspondence for GL 2 (Q p ) [7]. Existing proofs of the Cherbonnier Colmez theorem, including a generalization to families of representations by Berger and Colmez [3], rely on some calculations involving a formalism for decompletion in continuous Galois cohomology, inspired by results of Tate and Sen and later axiomatized by Colmez. However, one can express the proof in such a way that one makes essentially the same calculations on (ϕ, Ŵ)-modules as in [5], but without any need to introduce the Tate-Sen formalism. Besides making the proof more transparent, this approach gives rise to analogous results for representations of the étale fundamental groups of some rigid analytic spaces; for instance, the theory of overconvergent relative (ϕ, Ŵ)-modules introduced by Andreatta and Brinon [1] is generalized in [19] using this approach. (It should similarly be possible to recover the results of [3] in this fashion, and even to obtain a common generalization with [19].) 1 Comparison of Galois groups 1.1 Preliminaries on strict p rings We begin by recalling some basic properties of strict p-rings underlying the constructions made later, following the derivations in [27, 5]. (All rings considered will be commutative and unital.) Lemma For any ring R and any nonnegative integer n, the map x x pn induces a well-defined multiplicative monoid map θ n : R/(p) R/(p n+1 ). Proof If x y (mod p m ) for some positive integer m, then x p y p = (x y)(x p 1 + +y p 1 ) and the latter factor is congruent to px p 1 modulo p m ; hence x p y p (mod p m+1 ). This proves that θ n is well-defined; it is clear that it is multiplicative.

3 Page 3 of 31 Definition A ring R of characteristic p is perfect if the Frobenius homomorphism x x p is a bijection; this forces R to be reduced. (If R is a field, then R is perfect if and only if every finite extension of R is separable.) A strict p-ring is a p-torsion-free, p-adically complete ring R for which R/(p) is perfect, regarded as a topological ring using the p-adic topology. Example The ring Z p is a strict p-ring with Z p /(p) = F p. Similarly, for any (possibly infinite) set X, if we write Z[X] for the polynomial ring over Z generated by X and put Z[X p ]= n=0 Z[Xp n ], then the p-adic completion R of Z[X p ] is a strict p-ring with R/(p) = F p [X p ]. Lemma Let R be a perfect ring of characteristic p, let S be a p-adically complete ring, and let π : S S/(p) be the natural projection. Let t : R S/(p) be a ring homomorphism. Then there exists a unique multiplicative map t : R S with π t = t. In fact, t(x) xn pn (mod p n+1 ) for any nonnegative integer n and any x n S lifting t(x p n ). Proof This is immediate from Lemma Definition Let R be a strict p-ring. By the case S = R of Lemma 1.1.4, the projection R R/(p) admits a unique multiplicative section [ ] : R/(p) R, called the Teichmüller map. (For example, the image of [ ] : F p Z p consists of 0 together with the (p 1)-st roots of unity in Z p.) Each x R admits a unique representation as a p-adically convergent sum n=0 p n [x n ] for some elements x n R/(p), called the Teichmüller coordinates of x. Lemma Let R be a strict p-ring, let S be a p-adically complete ring, and let π : S S/(p) be the natural projection. Let t : R/(p) S be a multiplicative map such that t = π t is a ring homomorphism. Then the formula ( ) T p n [x n ] = p n t(x n ) (x 0, x 1,... R/(p)) ( ) n=0 n=0 defines a (necessarily unique) p-adically continuous homomorphism T : R S such that T [ ] = t. Proof We check by induction that for each positive integer n, T induces an additive map R/(p n ) S/(p n ). This holds for n = 1 because π t is a homomorphism. Suppose the claim holds for some n 1. For x =[x]+px 1, y =[y]+py 1, z =[z]+pz 1 R with x + y = z, [z] ([x p n ]+[y p n ]) pn (mod p n+1 ) t(z) (t(x p n ) + t(y p n )) pn (mod p n+1 ) by Lemma In particular,

4 Page 4 of 31 p n 1 ( ) p n T([z]) T([x]) T([y]) t(x i ip n y 1 ip n ) (mod p n+1 ). ( ) i=1 ( ) On the other hand, since p 1 p n Z for i = 1,..., p i n 1, we may write z 1 x 1 y 1 = [x]+[y] [z] p n 1 ( ) 1 p n [x p p i ip n y 1 ip n ] (mod p n ), i=1 apply T, invoke the induction hypothesis on both sides, and multiply by p to obtain pt(z 1 ) pt(x 1 ) pt(y 1 ) p n 1 i=1 ( p n i ) t(x ip n y 1 ip n ) (mod p n+1 ). ( ) Since T(x) = T([x]) + pt(x 1 ) and so on, we may add ( ) and ( ) to deduce that T(z) T(x) T(y) 0 (mod p n+1 ), completing the induction. Hence T is additive; it is also clear that T is p-adically continuous. From this it follows formally that T is multiplicative: for x = n=0 p n [x n ], y = n=0 p n [y n ], T(x)T(y) = p m+n t(x m )t(y n ) = p m+n t(x m y n ) m,n=0 = T m,n=0 m,n=0 We conclude that T is a ring homomorphism as claimed. p m+n [x m y n ] = T(xy). Remark Take R as in Example with X ={x, y}. By Lemma 1.1.4, we have [x] [y] = p n [P n (x, y)] n=0 ( ) for some P n (x, y) in the ideal (x p, y p ) F p [x p, y p ] and homogeneous of degree 1. By Lemma 1.1.6, ( ) is also valid for any strict p-ring R and any x, y R/(p). One can similarly derive formulas for arithmetic in a strict p-ring in terms of Teichmüller coordinates; these can also be obtained using Witt vectors (Definition 1.1.9). Theorem The functor R R/(p) from strict p-rings to perfect rings of characteristic p is an equivalence of categories. Proof Full faithfulness follows from Lemma To prove essential surjectivity, let R be a perfect ring of characteristic p, choose a surjection ψ : F p [X p ] R for some set X, and put I = ker(ψ). Let R 0 be the p-adic completion of Z[X p ]; as in Example 1.1.3, this is a strict p-ring with R 0 /(p) = F p [X p ]. Put I ={ n=0 p n [x n ] R 0 : x 0, x 1,... I}; this forms an ideal in R 0 by Remark Then R = R 0 /I is a strict p-ring with R/(p) = R.

5 Page 5 of 31 Definition For R a perfect ring of characteristic p, we write W (R) for the unique (by Theorem 1.1.8) strict p-ring with W (R)/(p) = R. This is meant as a reminder that W (R) also occurs as the ring of p-typical Witt vectors over R; that construction obtains the formulas for arithmetic in Teichmüller coordinates in an elegant manner linked to symmetric functions. Remark Let R be a strict p-ring. Since R is p-adically complete, the Jacobson radical of R contains p. In particular, if R is local, then so is R. 1.2 Perfect norm fields Theorem is obtained by matching up the Galois correspondences of the fields F p ((π)) and Q p (µ p ). The approach taken by Fontaine and Wintenberger is to pass from characteristic 0 to characteristic p by looking at certain sequences of elements of finite extensions of Q p in which each term is obtained from the succeeding term by taking a certain norm between fields; the resulting functor is thus commonly called the functor of norm fields. It is here that some careful analysis of higher ramification theory is needed in order to make the construction work. While the Fontaine Wintenberger construction gives rise directly to finite extensions of F p ((π)), it was later observed that a simpler construction (used repeatedly by Fontaine in his further study of p-adic Hodge theory) could be used to obtain the perfect closures of these finite extensions. Originally the construction of these perfect norm fields depended crucially on the prior construction of the imperfect norm fields of Fontaine Wintenberger (as in the exposition in [4]), but we will instead work directly with the perfect norm fields. Definition By an analytic field, we will mean a field which is complete with respect to a multiplicative nonarchimedean norm. For an analytic field, put o ={x : x 1}; this is a local ring with maximal ideal m ={x : x < 1}. We say has mixed characteristics if p = p 1 (so has characteristic 0) and the residue field κ = o /m of has characteristic p. Remark Any finite extension L of an analytic field is itself an analytic field; that is, the norm extends uniquely to a multiplicative norm on the extension field. A key fact about this extension is rasner s lemma: if P(T) [T] splits over L as n i=1 (T α i ) and β satisfies α 1 β < α 1 α i for i = 2,..., n, then α 1. A crucial consequence of rasner s lemma is that an infinite algebraic extension of is separably closed if and only if its completion is algebraically closed. More precisely, rasner s lemma is only needed for the if implication; the only if implication follows from the fact that the roots of a polynomial vary continuously in the coefficients. This principle also appears in the proof of Lemma Remark Using rasner s lemma, one may show that Q p (µ p ) and its completion have the same Galois group. Similarly, F p ((π)), its perfect closure, and the completion of

6 Page 6 of 31 the perfect closure all have the same Galois group. Consequently, from the point of view of proving Theorem 0.0.1, there is no harm in considering only analytic fields. Definition Let be an analytic field of mixed characteristics. Let o be the inverse limit lim o /(p) under Frobenius. In symbols, { } o = (x n ) o /(p) : x p n+1 = x n. n=0 By construction, this is a perfect ring of characteristic p: the inverse of Frobenius is the shift map (x n ) n=0 (x n+1) n=0. By applying Lemmas and to the homomorphism θ : o o /(p), we obtain a multiplicative map θ : o o and a homomorphism : W (o ) o. For x = (x 0, x 1,...) o, define x = θ(x). If we lift x n o /(p) to x n o, then x = x n pn whenever x n > p. Consequently, is a multiplicative nonarchimedean norm on o under which o is complete (given any Cauchy sequence, the terms in any particular position eventually stabilize). Lemma With notation as in Definition 1.2.4, for x, y o, x is divisible by y if and only if x y. Proof If x is divisible by y, then x y x/y y. Conversely, suppose x y. If y = 0, then x = 0 also and there is nothing more to check. Otherwise, write x = (x 0, x 1,...), y = (y 0, y 1,...), and choose lifts x n, y n of x n, y n to o. Since y = 0, we can find an integer n 0 0 such that yn p 1+1/p for n = n 0, and hence also for n n 0. Then for n n 0, the elements z n = x n /y n o have the property that p z n+1 z n p 1/p. By writing z p2 n+2 = (z n+1 + (z p n+2 z n+1)) p, we deduce that z p2 n+2 zp n+1 p 1. We thus produce an element z = (z 0, z 1,...) with x = yz by taking z n to be the reduction of z p n+1 for n n Definition eep notation as in Definition By Lemma 1.2.5, o is the valuation ring in an analytic field which is perfect of characteristic p. We call the perfect norm field associated to. (Scholze [21] calls the tilt of and denotes it by.) Exercise The formula x (θ(x p n )) n=0 defines a multiplicative bijection from o to the inverse limit of multiplicative monoids (but not of rings) { } (x n ) o : x p n+1 = x n. n=0 This map extends to a multiplicative bijection from to the inverse limit of multiplicative monoids { } (x n ) : x p n+1 = x n. n=0

7 Page 7 of Perfectoid fields In general, the perfect norm field functor is far from being faithful. Exercise Let be a discretely valued analytic field of mixed characteristics. Then is isomorphic to the maximal perfect subfield of κ. To get around this problem, we restrict attention to analytic fields with a great deal of ramification. (The term perfectoid is due to Scholze [21].) Definition An analytic field is perfectoid if is of mixed characteristics, is not discretely valued, and the p-th power Frobenius endomorphism on o /(p) is surjective. The following statements imply that taking the perfect norm field of a perfectoid analytic field does not concede too much information. Lemma Let be a perfectoid analytic field. (a) We have = ( ). (b) The projection o lim o /(p) o /(p) is surjective and induces an isomorphism o /(z) = o /(p) for any z o with z = p 1 (which exists by (a)). In particular, we obtain a natural isomorphism κ = κ. (c) The map : W (o ) o is also surjective. (The kernel of turns out to be a principal ideal; see Corollary ) Proof Since is not discretely valued, we can find r such that p 1 r p (p 1,1); the surjectivity of Frobenius then implies p 1/p r. This implies p 1/p, so p 1/p ( ) and p 1 ( ). Since also (p 1,1) = ( ) (p 1,1), we obtain (a), and (b) and (c) follow easily. Corollary The perfect norm field functor on perfectoid analytic fields is faithful. Example Take to be the completion of Q p (µ p ) for the unique extension of the p-adic norm, and fix a choice of a sequence {ζ p n} n=0 in which ζ p n is a primitive pn -th root of unity and ζ p p n+1 = ζ p n. The field is perfectoid because o /(p) = F p [ζ p, ζ p 2,...]/(1 + ζ p + +ζ p 1 p, ζ p ζ p p 2,...). By the same calculation, we identify with the completed perfect closure of F p ((π)) by identifying π with (ζ 1 1, ζ p 1,...). This example underlies the theory of (ϕ, Ŵ)-modules; see Sect. 2. Example Let F be a discretely valued analytic field of mixed characteristics (e.g., a finite extension of Q p ) and choose a uniformizer π of F. Choose a sequence π 0, π 1,... of elements of an algebraic closure of F in which π 0 = π and π p n+1 = π n for n 0. Take to be the completion of F(π 0, π 1,...); then o is the completion of o F [π 1, π 2,...], so

8 Page 8 of 31 o /(π) = F p [π 1, π 2,...]/(π p 1, π 1 π p 2,...). By Exercise below, is perfectoid. By the same calculation, we identify with the completed perfect closure of κ F ((π)) by identifying π with (π 0, π 1,...). This example underlies the theory of Breuil isin modules, which provide a good replacement for (ϕ, Ŵ)-modules for the study of crystalline representations; see [8, 11]. Let be an analytic field of mixed characteristics which is not dis- Exercise cretely valued. (a) Assume that there exists ξ with p 1 ξ < 1 such that Frobenius is surjective on o /(ξ). Then is perfectoid. (Hint: first imitate the proof of Lemma 1.3.3(a) to construct an element of with norm ξ 1/p, then construct pth roots modulo successively higher powers of ξ.) (b) Suppose that there exists an ideal I m such that the I-adic topology and the norm topology on o coincide, and Frobenius is surjective on o /I. Then is perfectoid. (Note that I need not be principal; for instance, take I ={x o : x < p 1/2 }. If I is finitely generated, however, it is principal.) Exercise An analytic field of mixed characteristics is perfectoid if and only if for every x, there exists y with y p x p 1 x. As in the previous exercise, one can also replace p 1 by any constant value in the range [p 1,1). Remark When developing the theory of norm fields, it is typical to consider the class of arithmetically profinite algebraic extensions of Q p, i.e., those for which the Galois closure has the property that its higher ramification subgroups are open. One then shows using Exercise that the completions of such extensions are perfectoid. While this construction has the useful feature of providing many examples of perfectoid fields, we will have no further need for it here. (See however Remark ) 1.4 Inverting the perfect norm field functor So far, we have a functor from perfectoid analytic fields to perfect analytic fields of characteristic p. In order to invert this functor, we must also keep track of the kernel of the map ; this kernel turns out to contain elements which behave a bit like linear polynomials, in that they admit an analogue of the division algorithm. This exploits an imperfect but useful analogy between strict p-rings and rings of formal power series, in which p plays the role of a series variable and the Teichmüller coordinates (Definition 1.1.5) play the role of coefficients. For a bit more on this analogy, see Remark ; for further discussion, including a form of Weierstrass preparation in strict p-rings, see for instance [8]. Hypothesis Throughout this section, let F be an analytic field which is perfect of characteristic p. We denote the norm on F by. Remark We will use frequently the following consequence of the homogeneity aspect of Remark 1.1.7: the function

9 Page 9 of 31 n=0 p n [x n ] sup{ x n } n satisfies the strong triangle inequality, and hence defines a norm on W (o F ). See Lemma for a related observation. Definition For z W (o F ) with reduction z o F, we say z is primitive if z = p 1 and p 1 (z [z]) W (o F ). (The existence of such an element in particular forces the norm on F not to be trivial.) Since W (o F ) is a local ring by Remark , for z = n=0 p n [z n ] it is equivalent to require that z 0 = p 1 and z 1 = 1; in particular, whether or not z is primitive depends only on z modulo p 2. Exercise If z W (o F ) is primitive, then so is uz for any u W (o F ). (Hint: for y = uz, show that y1 u 0 z 1 < 1 by working with Witt vectors modulo p 2.) In order to state the division lemma for primitive elements, we need a slightly wider class of elements of W (o F ) than the Teichmüller lifts. Definition An element x = n=0 p n [x n ] W (o F ) is stable if x n x 0 for all n > 0. Note that 0 is stable under this definition. An element of W (o F ) is stable if and only if it equals a unit times a Teich- Lemma müller lift. Proof This is immediate from the fact that [x] n=0 p n [y n ]= n=0 p n [xy n ]. Here is the desired analogue of the division lemma, taken from [16, Lemma 5.5]. Lemma For any primitive z W (o F ), every class in W (o F )/(z) is represented by a stable element of W (o F ). Proof Write z =[z]+pz 1 with z 1 W (o F ). Given x W (o F ), put x 0 = x. Given x l = n=0 p n [x l,n ] congruent to x modulo z, put x l,1 = n=0 p n [x l,n+1 ] and x l+1 = x l x l,1 z1 1 z, so that x l+1 is also congruent to x modulo z. Suppose that for some l, we have xl,n < p xl,0 for all n > 0. By Remark and the equality x l+1 =[x l,0 ] x l,1 z1 1 [z], we have xl+1,n xl,0 for all n 0. Also, xl+1,0 equals x l,0 plus something of lesser norm (namely zx l,1 times the reduction of z1 1 ), so xl+1,0 = xl,0. Hence xl+1 is a stable representative of the congruence class of x. We may thus suppose that no such l exists. By Remark again, sup n { xl+1,n } p 1 sup n { xl,n } for all l. The sum y = l=0 x l,1 z1 1 thus converges for the (p, [z])-adic topology on W (o F ) and satisfies x = yz; that is, 0 is a stable representative of the congruence class of x. Remark One might hope that one can always take the stable representative in Lemma to be a Teichmüller lift, but this is in general impossible unless the field F is not only complete but also spherically complete. This condition means that any

10 Page 10 of 31 decreasing sequence of balls in F has nonempty intersection; completeness only imposes this condition when the radii of the balls tend to 0. For example, a completed algebraic closure of Q p, or of a power series field in characteristic p, is not spherically complete. Lemma Any stable element of W (o F ) divisible by a primitive element must equal Proof Suppose x W (o F ) is stable and is divisible by a primitive element z. Put y = x/z and write x = n=0 p n [x n ], y = n=0 p n [y n ]. Also write z =[z]+pz 1 with z 1 W (o F ). Then on one hand, x = y 0 z, so y0 = p x0. On the other hand, by writing x = pz 1 y +[z]y and using Remark 1.4.2, we see that (because the term pz 1 y dominates) sup{ x n }=sup{ y n }. n n Since x is stable, this gives a contradiction unless x = 0, as desired. Corollary Suppose that z W (o F ) is primitive and that x, y W (o F ) are stable and congruent modulo z. Then the reductions of x, y modulo p have the same norm. Proof Put w = x y and write w = n=0 p n [w n ], x = n=0 p n [x n ], y = n=0 p n [y n ]. By Remark 1.4.2, w n max{ x 0, y0 } for all n 0. However, if x0 = y0, then w 0 = max{ x 0, y0 } > 0, so w is a nonzero stable element of W (of ) divisible by z. This contradicts Lemma 1.4.9, so we must have x 0 = y0 as desired. Exercise Give another proof of Lemma by formulating a theory of Newton polygons for elements of W (o F ). Remark A good way to understand the preceding discussion is to compare it to the theory of Weierstrass preparation for power series over a complete discrete valuation ring. For a concrete example, consider the ideal (T p) in the ring Z p [[T]]. There is a natural map Z p Z p [[T]]/(T p); one may see that this map is injective by observing that no nonzero element of Z p can be divisible by T p (by analogy with Lemma 1.4.9), and that it is surjective by observing that one can perform the division algorithm on power series to reduce them modulo T p to elements of Z p (by analogy with Lemma 1.4.7). In the situation considered here, however, we do not start with a candidate for the quotient ring W (o F )/(z). Instead, we must be a bit more careful in order to read off the properties of the quotient directly from the division algorithm. We are now ready to invert the perfect norm field functor. Theorem Choose any primitive z W (o F ) and put o = W (o F )/(z). For x o, apply Lemma to find a stable element y = n=0 p n [y n ] W (o F ) lifting x, then define x = y0. This is independent of the choice of y thanks to Lemma

11 Page 11 of 31 (a) The function is a multiplicative norm on o under which o is complete. (b) There is a natural (in F) isomorphism o /(p) = o F /(z). (c) The ring o is the valuation ring of an analytic field of mixed characteristics. (d) The field is perfectoid and there is a natural isomorphism = F. (e) The kernel of : W (o F ) o is generated by z. Proof Part (a) follows from the fact that the product of stable elements is stable (thanks to Remark 1.4.2). Part (b) follows by comparing both sides to W (o F )/(p, [z]). Part (c) follows from the fact (a consequence of Lemma 1.4.6) that if x = n=0 p n [x n ], y = n=0 p n [y n ] are stable and x 0 y0, then x is divisible by y in W (of ). Part (d) follows from (b) plus Theorem Part (e) follows from the construction of, or more precisely from the fact that every nonzero class in W (o F )/(z) has a nonzero stable representative. Corollary Let be a perfectoid analytic field. Then there exists a primitive element z in the kernel of : W (o ) o, so ker( ) is principal generated by z by Theorem (Exercise then implies that conversely, any generator of ker( ) is primitive.) Proof Since and have the same norm group by Lemma 1.3.3, we can find z o with z = p 1. Then θ(z) is divisible by p in o ; since is surjective, we can find z 1 W (o ) with (z 1 ) = θ(z)/p. This forces z 1 W (o ), as otherwise we would have (z 1 ) < 1. Now z =[z]+pz 1 is a primitive element of ker( ), as desired. Example In Example 1.3.5, note that π = p p/(p 1). One may then check that z = ([1 + π] 1)/([1 + π] 1/p 1) = [1 + π] i/p is a primitive element of W (o ) belonging to ker( ). Hence z generates the kernel by Theorem Example One can also write down explicit primitive elements in some cases of Example A simple example is when π = p (this forces the field F to be absolutely unramified). In this case, z = p [π] is a primitive element of W (o ) belonging to (and hence generating) ker( ). p 1 i=0 1.5 Compatibility with finite extensions At this point, using Theorem and Corollary , we obtain the following statement, which one might call the perfectoid correspondence (or the tilting correspondence in the terminology of [21]).

12 Page 12 of 31 Theorem (Perfectoid correspondence) The operations (, ker( : W (o ) o )), (F, I) W (o F )[p 1 ]/I define an equivalence of categories between perfectoid analytic fields and pairs (F, I) in which F is a perfect analytic field of characteristic p and I is an ideal of W (o F ) generated by a primitive element. Remark From Examples and 1.3.6, we see that one cannot drop the ideal I in Theorem : one can have nonisomorphic perfectoid analytic fields whose perfect norm fields are isomorphic. We will establish that the perfectoid correspondence is compatible with finite extensions of fields on both sides, where on the right side we replace the ideal I by its extension to the larger ring. This will give Theorem by taking to be the completion of Q p (µ p ). However, the more general result for an arbitrary perfectoid is relevant for extending p-adic Hodge theory to a relative setting; see Remark The first step is to lift finite extensions from characteristic p; this turns out to be straightforward. Lemma Let be a perfectoid analytic field and put I = ker( : W (o ) o ). Let F be a finite extension of and put L = W (o F )[p 1 ]/IW (o F )[p 1 ]. Then [L : ]=[F : ]. (Note that by Theorem , L is a perfectoid analytic field and we may identify L with F.) Proof Apply Corollary to choose a primitive generator z I. The extension F/ is separable because is perfect; it thus has a Galois closure F. Put L = W (o F )[p 1 ]/(z), which is a perfectoid analytic field thanks to Theorem For any subgroup H of G = Gal( F/ ), averaging over H defines a projection L = W (o F )[p 1 ] zw (o F )[p 1 ] W (o F H )[p 1 ] zw (o F )[p 1 ] W (o F H )[p 1 ] = W (o F H )[p 1 ] zw (o F H )[p 1 ], so the right side equals the fixed field L H. Put G = Gal( F/F). Apply the above analysis with H = G and H = G; since F G = and F G = F, we obtain L G = and L G = L. The action of G on L is faithful: any nontrivial g G moves some x F, and for n sufficiently large we may write θ([x p n ]) g(θ([x p n ])) θ([(x g(x)) p n ]) p 1 x p n to see that g also moves θ([x p n ]). By Artin s lemma, L is a finite Galois extension of L H with Galois group H, so [L : ]=[ L : ]/[ L : L] =#G/# G =[ F : ]/[ F : F] =[F : ], as desired.

13 Page 13 of 31 We still need to check that a finite extension of a perfectoid analytic field is again perfectoid. It suffices to study the case where the perfect norm field is algebraically closed. It is this argument in particular that we learned from the course of Coleman mentioned in the introduction, and even wrote down on one previous occasion; see [13, Theorem 7]. Lemma If is a perfectoid field and is algebraically closed, then so is. Proof Let P(T) o [T] be an arbitrary monic polynomial of degree d 1; it suffices to check that P(T) has a root in o. We will achieve this by exhibiting a sequence x 0, x 1,... of elements of o such that for all n 0, P(x n ) p n and x n+1 x n p n/d. This sequence will then have a limit x o which is a root of P. To begin, take x 0 = 0. Given x n o with P(x n ) p n, write P(T + x n ) = i Q it i. If Q 0 = 0, we may take x n+1 = x n, so assume hereafter that Q 0 = 0. Put c = min{ Q0 /Q j 1/j : j > 0, Qj = 0}; by taking j = d, we see that c Q 0 1/d. Also, since has the same norm group as by Theorem , this norm group is divisible; we thus have c = u for some u o. (Note that c is defined so as to equal the smallest norm of a norm of Q in, but this fact is not explicitly used in the proof.) Apply Corollary to construct a primitive element z ker( ). Put R 0 = 0. For each i > 0, choose R i o whose image in o /(z) = o /(p) is the same as that of Q i u i /Q 0. Define the (not necessarily monic) polynomial R(T) = i R it i o [T]. By construction, the largest slope in the Newton polygon of R is 0; by this observation plus the fact that is algebraically closed, it follows that R(T) has a root y o. Choose y o whose image in o /(p) = o /(z) is the same as that of y, and take x n+1 = x n + uy. Then i Q iu i y i /Q 0 0 (mod p), so P(x n+1 ) p 1 Q 0 p n 1 and x n+1 x n = u Q 0 1/d p n/d. We thus obtain the desired sequence, proving the claim. Remark The inertia subgroup of the Galois group of a finite extension of analytic fields is solvable; see for instance [14, Chapter 3] and references therein. (The discretely valued case may be more familiar; for that, see also [27, Chapter IV].) Using this statement, one can give a slightly simpler proof of Lemma by considering only cyclic extensions of prime degree. We chose not to proceed this way so as to make good on our promise to keep the proof of Theorem entirely free of ramification theory. We are now ready to complete the proof of Theorem Theorem Let be a perfectoid analytic field. Then every finite extension of is perfectoid, and the operation L L defines a functorial correspondence between the finite extensions of and. In particular, the absolute Galois groups of and are homeomorphic. Proof Apply Corollary to construct a primitive generator z ker( : W (o ) o ). Let M be the completion of an algebraic closure of ; it is again

14 Page 14 of 31 algebraically closed by Remark By Theorem 1.5.1, M arises as the perfect norm field of a perfectoid analytic field M, which by Lemma is also algebraically closed. By Lemma 1.5.3, each finite Galois extension L of within M is the perfect norm field of a finite Galois extension L of within M which is perfectoid. The union L of such fields L is dense in M because the union of the L is dense in M. By Remark 1.2.2, L is algebraically closed; that is, every finite extension of is contained in a finite Galois extension which is perfectoid. The rest follows from Theorem Remark The proof of Theorem is a digested version of the one given in [18, Theorem 3.5.6]. A different proof has been given by Scholze [21, Theorem 3.7], in which the analysis of strict p-rings is supplanted by use of a small amount of almost ring theory, as introduced by Faltings and developed systematically by Gabber and Ramero [11]. However, these two approaches resemble each other far more strongly than either one resembles the original arguments of Fontaine and Wintenberger. Remark In both [18, 21], Theorem is generalized to a statement relating the étale sites of certain nonarchimedean analytic spaces in characteristic 0 and characteristic p, including an optimally general form of Faltings s almost purity theorem. (For the flavor of this result, see Theorem ) This is used as a basis for relative p-adic Hodge theory in [18, 19, 22]. Note that for this application, it is crucial to have Theorem and not just Theorem 0.0.1: one must use analytic spaces in the sense of Huber (adic spaces) rather than rigid analytic spaces, which forces an encounter with general analytic fields. One must also deal with valuations of rank greater than 1 (i.e., whose value groups do not fit inside the real numbers), but this adds no essential difficulty. The following is taken from [18, Proposition 3.5.9]. Exercise Let L / be a finite extension of analytic fields such that L is perfectoid. Prove that is also perfectoid. Hint: reduce to the Galois case, then produce a Galoisinvariant primitive element. 1.6 Some applications We describe now a couple of applications of Theorem 1.5.6, in which one derives information in characteristic 0 by exploiting Frobenius as if one were working in positive characteristic. Definition An analytic field is deeply ramified if for any finite extension L of, ol /o = 0; that is, the morphism Spec(o L ) Spec(o ) is formally unramified. (Beware that this morphism is usually not of finite type if is not discretely valued.) Theorem Any perfectoid analytic field is deeply ramified. Proof Let be a perfectoid field and let L be a finite extension of. Choose x 1,..., x n o L which form a basis of L over ; we can then find t o {0} such that to L o x 1 + +o x n. Since L is a finite separable extension of, L/ = 0;

15 Page 15 of 31 consequently, we can choose u o {0} so that u dx i vanishes in ol /o for each i. For any x o L, t dx = d(tx) is a o -linear combination of x 1,..., x n, so tu dx = 0. On the other hand, L is perfectoid by Theorem Hence for any x o L, we can find y o L for which x y p (mod p); this implies that ol /o = p ol /o. As a result, ol /o = p n ol /o for any positive integer n; by choosing n large enough that tu is divisible by p n, we deduce that ol /o = 0. Hence is deeply ramified. Remark Theorem admits the following converse: any analytic field of mixed characteristics which is deeply ramified is also perfectoid. See [11, Proposition 6.6.6]. Theorem Let be a perfectoid analytic field and let L be a finite extension of. Then Trace : m L m is surjective. Proof By Theorem 1.5.6, L is also perfectoid. Let, L be the perfect norm fields of, L. Since L is a finite separable extension of, there exists u m such that uo Trace(m L ). By applying the inverse of Frobenius, we obtain the same conclusion with u replaced by u p n for each positive integer n. Hence Trace : m L m is surjective. Since is not discretely valued, we can find t m with p 1 < t < 1. Since L is a finite separable extension of, there exists a nonnegative integer m such that (p/t) m m Trace(m L ). If m > 0, then for each x (p/t) m 1 m, by the previous paragraph we may write x = Trace(y) + pz for some y m L, z o ; since pz = (p/t)(tz) (p/t) m m, it follows that z Trace(m L ) and hence x Trace(m L ). In other words, we may replace m by m 1; this proves the desired result. Remark Note that Theorem still holds, with the same proof, if we replace the perfectoid field by a dense subfield as long as the norm extends uniquely to the finite extension L of (i.e., o is henselian in the sense of Lemma below). For instance, we may take to be an infinite algebraic extension of an analytic field F of mixed characteristics. One important case is when F = Q p and is a Galois extension whose Galois group contains Z p (e.g., any p-adic Lie group). In this case, the perfectoid property can be checked using a study of higher ramification groups; the technique was introduced by Tate and developed further by Sen [26] (see also [16, 13]). However, no ramification theory is necessary in case is a field for which the perfectoid property can be checked directly (as in Examples or 1.3.6), or a finite extension of such a field (using Theorem 1.5.6). 1.7 Gauss norms We conclude this section by appending some more observations about norms on strict p-rings, for use in the second half of the paper. Hypothesis Throughout Sect. 1.7, again let F be an analytic field which is perfect of characteristic p, with norm.

16 Page 16 of 31 The key construction here is an analogue of the Gauss norm, following [16, Lemma 4.1]. Lemma For r > 0, the formula p n [x n ] = sup{p n ( x n ) r } r n=0 n ( ) defines a function r : W (F) [0, + ] satisfying the strong triangle inequality x + y r max{ x r, y r } and the multiplicativity property xy r = x r y r for all x, y W (F) (under the convention 0 + = 0). In particular, the subset W r (F) of W (F) on which r is finite forms a ring on which r defines a multiplicative nonarchimedean norm. Proof From Remark 1.1.7, it follows that [x]+[y] r max{ [x] r, [y] r } for all x, y o F (as in Remark 1.4.2). It follows easily that for x, y F, x + y r max{ x r, y r } and xy r x r y r. To establish multiplicativity, by continuity it is enough to consider finite sums x = M m=0 p m [x m ], y = N n=0 p N [y n ]. For all but finitely many r > 0, the quantities p m n ( xm y n ) r (m = 0,..., M; n = 0,..., N) are all distinct; for such r, the fact that r is a nonarchimedean norm (shown above) gives xy r = max{p m n ( xm y n ) r : m = 0,..., M; n = 0,..., N} = x r y r. Since x r, y r, xy r are all continuous functions of r, we may infer multiplicativity also in the exceptional cases. Remark For s (0, r] and x = n=0 p n [x n ], x s = sup{p n ( x n ) s } sup{p ns/r ( x n ) s }= x s/r n n ( ) An even stronger statement is that for x W (F) fixed, the function log( r ) : W (F) R {+ } is convex (because it is the supremum of convex functions); this bears a certain formal resemblance to the Hadamard three circles theorem in complex analysis. A related observation is that r. { limr 0 + x r = 1 (x 0 = 0) lim sup r 0 + x r p 1 (x 0 = 0). ( ) Lemma For r > 0, the ring W r (F) is complete with respect to r. Proof For x = n=0 p n [x n ], y = n=0 p n [y n ] with x, y c, x y ǫc for some c > 0, ǫ [0, 1), Remark implies that xn y n cp n/r max{p m/r ǫ 1/pm : m = 0,..., n}. ( )

17 Page 17 of 31 Let x 1, x 2,... be a Cauchy sequence in W r (F), and write x i = n=0 p n [x i,n ]. Choose c such that x i r c for all i. Using ( ), we see that for each n, the sequence x i,n is Cauchy and hence converges to a limit x n F. Put x = n=0 p n [x n ]; we see easily that x r c and hence x W r (F). It thus remains to check that x is a limit of x 1, x 2,...; this formally reduces to the case x = 0. For each δ >0, by definition there exists N > 0 such that xi x j r cδ for all i, j N. For each n, by ( ) we have xi,n x j,n c 1/r p n/r max{p m/r δ 1/(pm r) : m = 0,..., n}; since the sequence x i,n converges to 0 as i, we must also have xi,n c 1/r p n/r max{p m/r δ 1/(pm r) : m = 0,..., n} and so x i r c max{p m δ 1/pm : m = 0, 1,...}. As δ 0, this upper bound tends to 0; this proves that x i 0 as desired. Lemma The ring W (F) = r>0 W r (F) is a local ring which is not complete but is henselian; that is, every finite étale F-algebra lifts uniquely to a finite étale W (F) -algebra. Proof We first check that W (F) is local; it suffices to check that p belongs to the Jacobson radical, meaning that any x = n=0 p n [x n ] W (F) with x 0 = 1 is a unit. By ( ), for r > 0 sufficiently small we have x 1 r < 1. By Lemma 1.7.4, the geometric series i=0 (1 x) i converges in W r (F) to a multiplicative inverse of x. We next check that W (F) is henselian. As described in [18, Remark 1.2.7] (see also [11, Theorem 5.11] and [20, Exposé XI, 2]), to check that W (F) is henselian, it suffices to check a special form of Hensel s lemma: every monic polynomial P(T) = i P it i W (F)[T] with P 0 pw (F), P 1 / pw (F) has a root in pw (F). By the previous paragraph, P 1 W (F). By ( ), for r > 0 sufficiently small, P 0 /P 1 r < 1, P i (P 0 /P 1 ) i r < P 1 2 r, Pi (P 0 /P 1 ) i 1 r < P 1 r (i 2). In particular, for x 0 = P 0 /P 1, for r > 0 sufficiently small we have P (x 0 ) r < P(x 0 ) 2 r, so the usual Newton Raphson iteration x n+1 = x n P(x n) P (x n ) converges to a root x of P(T) in W r (F). Since the iteration also converges p-adically, we must also have x pw (F); this proves the claim.

18 Page 18 of 31 Exercise We describe here a common variant of W r (F). (a) For any z o F with z < 1, the ring W (o F )[[z] 1 ] consists of those elements of W(F) with bounded Teichmüller coordinates. (b) The completion of W (o F )[[z] 1 ] under r consists of those x = n=0 p n [x n ] for which lim n p n ( x n ) r = 0. In particular, these form a ring, denoted W r (F). (c) The ring W r (F) is contained in W r (F) and contains W s (F) for all s > r. Consequently, we also have W (F) = r>0 W r (F). (d) The ring W r (F) is complete with respect to r. The rings W r (F) arise naturally in the geometric interpretation of p-adic Hodge theory given by Fargues and Fontaine [8]; this interpretation is facilitated by some useful noetherian properties of these rings, as described in [17]. Remark Let ϕ denote the endomorphism of W(F) induced by the Frobenius map on F. For any r > 0, using the fact that W (F) = n=0 W p nr (F) = n=0 ϕ n (W r (F)), it can be shown that every finite étale F-algebra lifts uniquely to a finite étale W r (F) -algebra. In case F = for some deeply ramified analytic field of mixed characteristics, the map : W (o F ) o extends to a homomorphism W r (F) for any r > 1, and Theorem is equivalent to the statement that every finite étale -algebra lifts uniquely to W r (F). (If we replace W r (F) with the ring W r (F) of Exercise 1.7.6, we may also take r = 1 in this last statement.) 2 Galois representations and (ϕ, Ŵ) modules Hypothesis Throughout Sect. 2, let denote a finite extension of Q p, and write G for the absolute Galois group of. Convention When working with a matrix A over a ring carrying a norm, we will write A for the supremum of the norms of the entries of A (rather than the operator norm or spectral radius). Definition Let FÉt(R) denote the category of finite étale algebras over a ring R. For R a field, these are just the direct sums of finite separable field extensions of R. 2.1 Some period rings over Q p We will consider four different rings which can classify G -representations. We first introduce them all in the case = Q p. Definition Let L be the completed perfect closure of F p ((π)), and put à Qp = W (L). This defines a complete topological ring both for the p-adic topology and for the weak topology, under which a sequence converges if each sequence of Teichmüller coordinates converges in the norm topology on L. (The restriction of the weak topology to W (o L ) coincides with the (p, [π])-adic topology.) Let ϕ be the endomorphism of à Qp induced by the Frobenius map on L.

19 Page 19 of 31 Definition Let A Qp be the p-adic completion of Z((π)); it is a Cohen ring (a complete discrete valuation ring with maximal ideal (p)) with A Qp /(p) = F p ((π)). We identify A Qp with a subring of à Qp in such a way that π corresponds to [1 + π] 1. Note that ϕ then acts on A Qp as the Z p -linear substitution π (1 + π) p 1, and a sequence in A Qp converges for the weak topology if and only if its image in A Qp /(p n ) = (Z/p n Z)((π)) converges π-adically for each positive integer n. Definition Put à Q p = W (L); by Lemma 1.7.5, this is an incomplete but henselian local ring contained in W (L) = à Qp. Note that ϕ acts bijectively on à Q p. We equip à Q p with the p-adic and weak topologies by restriction from à Qp ; we also define the LF topology, in which a sequence converges if and only if it converges in some W r (L). (LF is an abbreviation for limit of Fréchet.) Definition Put A Q p = à Q p A Qp ; since Z p [π ± ] A Q p, A Q p is again a henselian local ring with residue field F p ((π)) on which ϕ acts. It inherits p-adic, weak, and LF topologies. For a more concrete description of A Q p, see Corollary Definition For γ Ŵ = Z p, let γ : A Q p A Qp be the Z p -linear substitution π (1 + π) γ 1, where (1 + π) γ is defined via its binomial expansion. The induced map on F p ((π)) extends to L and thus defines an action of Ŵ on à Qp ; Ŵ also acts on à Q p and A Q p. For {Ã, A, Ã, A }, the action of Ŵ on Qp is continuous (meaning that the action map Ŵ Qp Qp is continuous) for the weak topology and (when available) the LF topology. Exercise In Definition 2.1.5, the action of Ŵ on Qp is not continuous for the p-adic topology, even though the action of each individual element of Ŵ is a continuous map from Qp to itself. 2.2 Extensions of Q p We extend the definition of the period rings to finite extensions of Q p using a refinement of Theorem Theorem For {Ã, A, Ã, A }, the category FÉt(Q p ) is equivalent to the category of finite étale algebras over Qp admitting an extension of the action of Ŵ. Moreover, this equivalence is compatible with the base extensions among different choices of. Proof Put L = à Qp /(p). Via Remark 1.2.3, Theorem 1.5.6, and the fact that the local rings à Q p and A Q p are both henselian, we see that the categories FÉt(Q p (µ p )), FÉt(L), FÉt(F p ((π))), FÉt(à Qp ), FÉt(A Qp ), FÉt(à Q p ), FÉt(A Q p ) are all equivalent, compatibly with base extensions among different choices of. It thus suffices to consider =à in what follows. From the explicit description given in Example 1.3.5, we see that the map : W (o L ) Z p [µ p ] becomes Ŵ-equivariant if we identify Ŵ with Gal(Q p (µ p )/Q p )

20 Page 20 of 31 via the cyclotomic character. Consequently, for FÉt(Q p ), the object in FÉt(à Qp ) corresponding to Qp Q p (µ p ) carries an action of Ŵ. Conversely, suppose S FÉt(à Qp ) carries an action of Ŵ; then the corresponding object E of FÉt(Q p (µ p )) also carries an action of Ŵ. We may realize E as the base extension of a finite étale algebra E n over Q p (µ p n) for some nonnegative integer n; by Artin s lemma, E n is fixed by a subgroup of Ŵ of finite index, which is necessarily open. By Galois descent, E n descends to a finite étale algebra E over Q p, as desired. Definition Let Ã, A, Ã, A be the objects corresponding to via Theorem We may write à = W ( L), à = W ( L) for L running over the connected components of à /(p) (which correspond to the connected components of Qp Q p (µ p )). We may thus equip with a p-adic topology, a weak topology, and (for =Ã, A ) also an LF topology. Define the norm on à /(p) as the supremum over connected components; for r > 0, define r on à as the supremum over connected components. The actions of ϕ, Ŵ extend to ; the action of Ŵ is again continuous for the weak topology and (when available) the LF topology. Note that Ŵ acts transitively on the L. Exercise Each of the topologies on coincides with the one obtained by viewing as a finite free module over Qp and equipping the latter with the corresponding topology. Remark Each connected component L of A /(p) is a finite separable extension of F p ((π)), and hence is itself isomorphic to a power series field in some variable π L over some finite extension F q of F p. In general, there is no distinguished choice of π L. One has similar (and similarly undistinguished) descriptions of A and A ; see Lemma Exercise In Remark 2.2.4, F q coincides with the residue field of (µ p ). Note that this may not equal the residue field of. Lemma eep notation as in Remark Let R be the connected component of A with R/(p) = L, and choose π L R lifting π L. Then R is isomorphic to the p-adic completion of W (F q )((π L )). Proof It suffices to observe that the latter ring is indeed a finite étale algebra over A Qp whose reduction modulo p is isomorphic to L. Exercise In Lemma 2.2.6, the weak topology on R (obtained by restriction from A ) coincides with the weak topology on the p-adic completion of W (F q )((π L )) (in which as in Definition 2.1.2, a sequence converges if and only if it converges π L -adically modulo each power of p). Lemma eep notation as in Lemma 2.2.6, but assume further that π L R for R the connected component of A with R /(p) = L. Then there exists r 0 > 0 (depending on L and π L ) with the following properties.